In 1976 I was about to become a student in the math department at Moscow State University. As an IMO team member I was accepted without entrance exams, but all of my other classmates had to take the exams. There were four exams: written math, oral math, physics, and an essay.

The written math exam was the first, and here are the problems. I want my non-Russian readers to see if they notice anything peculiar about this exam. Can you explain what is peculiar, and what might be the hidden agenda?

Problem 1. Solve the equation

Problem 2. Solve the inequality

Problem 3. Consider a right triangle ABC with right angle C. Angle B is 30° and leg CA is equal to 1. Let D be the midpoint of the hypotenuse AB, so that CD is a median. Choose F on the segment BC so that the angle between the hypotenuse and the line DF is 15°. Find the area of CDF. Calculate its numeric value with 0.001 precision.

Problem 4. Three balls, two of which are the same size, are tangent to the plane P, as well as to each other. In addition, the base of a circular cone lies on the plane P, and its axis is perpendicular to the plane. All three balls touch the cone from the outside. Find the angle between a generatrix of the cone and the plane P, given that the triangle formed by the points of tangency of the balls and the plane has one angle equal to 150°.

Problem 5. Let r < s < t be real numbers. If you set y equal to any of the numbers r, s or t in the equation x2 − (9 − y)x + y2 − 9y + 15 = 0, then at least one of the other two numbers will be a root of the resulting quadratic equation. Prove that −1 < r < 1.

Let me describe some background to this exam. Applicants who solve fewer than two problems fail the exam and are immediately rejected. People who solve two or three problems are given 3 points. Four problems earn 4 points, and five problems earn 5 points.

If you still do not see the hidden agenda, here is another clue. People who get 5 points on the first exam and, in addition, have a gold medal from their high school (that means all As) are admitted right after the first exam. For the others, if they do not fail any of the exams, points are summed up with their GPAs to compute their scores. The so-called half-passing score is then calculated. Scores strictly higher than the half-passing score qualify applicants for admission. However, there are too many applicants for the available openings with at least the half-passing score. As a result only some people with exactly the half-passing score are accepted, at the discretion of the department.

Now my readers have enough information to figure out the hidden agenda behind that particular exam.

20 Comments

LSK:

Well, one thing I can definitely say about the exam: it’s very hard! I won math competitions in high school, and still do them in college, and I doubt I could solve even one of these problems easily. Was the agenda to keep non-IMO students out?

any idiot with math at that level could solve the simple problems. Therefore, nearly everyone with similar GPA will be at the half-passing score. the department gets to pick their favorite entrants from among all the boys who apply. as long as they pick a girl and someone noone likes, they can avoid litigation.

Austin:

C.S., I had the same thought, but I’m doubtful for two reasons:

1) GPA is pretty granular, assuming it doesn’t get rounded off. A student who takes 50-60 courses on the 4.0/3.7/3.3/etc. scale can wind up with somewhere between 15000 and 20000 different GPAs (back-of-the-envelope calculation). Of course you’ll have a mode at 4.0 (or whatever the perfect GPA is), so the frequency of that mode will be the main factor in determining the size of the half-passing group (the peloton?). I guess the frequency of 4.0’s could be high if (a) there is lots of grade inflation or (b) the applicant pool is extremely self-selecting.

2) I’m not sure which problems are the ‘simple’ ones! The first commenter thought none of them were. I can do #1,2,3 automatically, but #4 starts to give me a bit of a headache. It is clearly doable, but on the other hand I wouldn’t call it simple. I didn’t think about #5 long enough to see an angle of attack — is it fiendishly hard? If not, then it seems likely that the full range of non-failing scores (3,4,5) would be realized among reasonably strong applicants.

I agree with Chao, though: it is a bit odd that a single problem counts for as much as the difference between an A average and a B average (assuming GPAs are out of 4.0). And how the heck do you grade a proof (#5) without the possibility of partial credit? Granted that a proof is either valid or not valid, but a proof can be invalidated by a minor technical flaw and still show the intelligence and capability that one would presumably be looking for.

Lennie Friedlander:

I am not exactly a person to whom Tanya’s question was addressed: I grew up in the Soviet union, and I went through entrance exams, though that was long time ago.
Nevertheless, I am somewhat baffled by what a hidden agenda might be. If this is to make sure that all minimally qualified candidates pass (2 or 3 problems
are straightforward), and almost nobody gets 5 points, so the gold medalists lose any advantage that they could have; as a consequence the oral exams, which are much more subjective, become decisive, then I would not call this agenda hidden.

I found problems 1,2,3 quite easy or extremely easy (well at my age!). Problems 4 & 5….still can’t think them through. They are not straightforward. So maybe the university is picking up very selective people or pick only the ones who solve problem 4 & 5. They know everyone will solve the first three.

Austin:

OK, I tried #4 and… wow, I spoke too soon when I said it was “clearly doable.”

There’s something about this problem that bothers me. A right circular cone with its base on P is determined by four parameters: the vertex angle and the spatial coordinates of the vertex. Generically, touching a sphere costs one degree of freedom. We are under three such constraints. Thus the solution appears underdetermined.

Is there something about the specific balls in this problem which makes the solution unique? I did notice one curious detail: if you drop a perpendicular to P from the point where the two large balls touch, then it just happens that this line is also tangent to the small ball. Is 90° a valid answer? 😉

Aaron F.:

Because of a book I once came across in my college’s library, I think I have a pretty good idea what the hidden agenda might be. I’m really terrible at these kinds of problems, though, so I can’t tell what’s peculiar about the exam.

Maybe it’s just my weak grasp of complex analysis talking, but Problem 1 looks a bit sinister to me. I found one unambiguous solution pretty easily, but I’m worried there may be be other solutions that work or don’t work depending on where you put the branch cut. Other people seemed to find this problem easy, so could one of you maybe explain it to me? (You can use rot13 to keep from spoiling everyone else’s fun.)

Ilya:

Animesh,
This is a standard MSU exam problem (they try to fail the students and always arrange solutions that do not belong to the domain of definition). You must, first, specify the domain of definition, which in this case is determined by the inequalities 0<sin(-x)0, and only then solve the equation sin(-x)=sin(x/2)+sin(3x/2)=2sin((x/2+3x/2)/2)cos((x/2-3x/2)/2)=2sin(x)cos(x/2). Thus, cos(x/2)=-1/2, hence x=4pi/3+4kpi or 8pi/3+4kpi. However, the first series of solutions doesn’t belong to the domain of definition. Hence x=8pi/3+4kpi.

The second problem is also easy, but one must be careful about the signs. The third problem is very easy.

The difficult problems are the last two. In the last problem you have an ellipse symmetric with respect to x=y (i.e. x=y is one of its axis). Thus, if (x,y) is a solution then so is (y,x). Hence, there is a line x=a that intersects the ellipse at points (a,b) and (a,c), and {a,b,c}={r,s,t}. If you draw the ellipse, you see that there are two points of intersection if and only if -1<x<7. Furthermore, a<min{b,c} if x<1, if x=1 then a=b (contradiction), if x=7 then b=c (contradiction), and if 1<x<7 then either b<1 or c<1. In any (admissible) event -1<min{a,b,c}<1. I have no idea how kids recently graduated from a high school are supposed to solve this. I tried also to look at problem 4. There is a lot of symmetry in this problem (in particular the center of the base of the cone lies on the line passing through the point of tangency of the smaller sphere with P and perpendicular to the interval joining the two points of tangency of the equiradial spheres. I tried to compute the angle (taking the symmetries into account it is not difficult), but got no solutions (most likely there is a mistake in my computations).

As I grew up in Moscow, I suppose I know the (not very hidden) agenda, and will not comment on this.

Sasha:

I gave these problems to my graduate students and nobody solved even the first problem correctly. Most of the students never have seen such problems and do not learn how to consider all the cases when a calculator is not allowed.
When you say that the problems are easy, and if you are not the USAMO winner, then you are not aware of how wrong you are…